differentiation

[Bestie]

How would I explain why the gradient of tan (x) is always positive, I'm not sure how to show it algebratically... i know that the derivative of tan(x) is sec(x)^2 and when i draw the graph of sec(x)^2 the y values are ll greater than 0 thus the grdient must always be positive but how would i show this alegratically with working out?

Thank you in advance

[RKTR]

y=tan(x) dy/dx=sec^2 x =1/ cos^2 x

range of cos x =[-1,1] 
range of cos^2(x) =[0,1]
range of 1/cos^2(x) = [1, infinity)
therefore it is always positive

[Bestie]

oh yea how stupid of me thank you

also, can you please help me here:
how would i simplify (50/4)cos(x)^2 + (18/4)sin(x)^2 + 8 cause apparently it needs to equal 5?

[kinslayer]

oh yea how stupid of me thank you

also, can you please help me here:
how would i simplify (50/4)cos(x)^2 + (18/4)sin(x)^2 + 8 cause apparently it needs to equal 5?

This never equals 5...



Something missing from the question, perhaps?

[keltingmeith]

On the first question, if you're having trouble explaining something algebraically, don't be scared to do a quick sketch of a graph. They can be used in questions to explain things, eg. why the gradient of tan(x) is always positive.

[lzxnl]

Graphs don't really help here.
You can just say the square of a real number is always positive. sec^2 (x) is the square of a real number as sec(x) returns real values for real values of x.

[Bestie]

thank you

question:
how would i find the coordinates of the inflexion of y = sin(x) for x [0,2pi]
what i have done so far:
f'(x) = cos(x)
f''(x) = -sin(x)
f''(x) = 0 , x = pi/2 and 3pi/2

i'm lost what now?

thank you in advance

[RKTR]

f''(x)=0
-sin(x)=0
x=0,pi,2pi  i think you did f'(x)=0 instead?

sub the values back into y=sin(x)

the coordinates are (0,0),(pi,0),(2pi,0)

[Bestie]

but then why is the ans only (pi,0) ?
and yup i accidentally did f'(x) = 0
sorry

[Thorium]

but then why is the ans only (pi,0) ?
and yup i accidentally did f'(x) = 0
sorry

It has to be only (pi,0) because x is restricted to [0,2pi]. I accept that 0 and 2pi are still included, but the points of inflexions at those points are not shown conpletely if you plot them. In contrast, the point of inflexion at (pi,0) is drawn completely.

[Bestie]

oh... cause the ans used something about proving that f''(x) is a minimum for it to be a point of inflection, but shouldn't f''(x) = 0

[EspoirTron]

oh... cause the ans used something about proving that f''(x) is a minimum for it to be a point of inflection, but shouldn't f''(x) = 0

Use a concavity test for f''(x)=0. If concavity changes then it is an inflection point, but not necessarily a stationary point of inflection.

Edit: I was confused.

[Bestie]

im confused too, cause then there is no need to do f''(x) cause you know there is an inflection already???

[Thorium]

im confused too, cause then there is no need to do f''(x) cause you know there is an inflection already???

There is a point of inflexion at x if and only if f"(x)=0 at x. Since f"(x) is the derivative function of f'(x), then we can say that a point of inflexion in f(x) is a local min/max in the f'(x).

[Bestie]

oh ok i get it now! thank you very much XD